Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Percents
Glencoe Math Course 2 Volume 1 Chapter 2 Exercise 2.2 Solutions Page 111 Exercise 1 Problem 1
We need to explain how we can percent help you understand situations
The percentage helps to understand situations involving money
The interest rates are written as percent
Also, find the interest earned on a savings account and the amount of interest charged on bank loans and credit cards.
The sales tax is also indicated in percents.
Hence explained.
Page 111 Exercise 2 Problem 2
Given:
About how many people took lessons at school?
To find – The number of people took lessons at school.
Total number of people surveyed = 200
Number of people took lessons at school = \(\frac{3}{10}\) 0f 200
\(\frac{3}{10}\) × 200 = 60
The number of people took lessons at school = 60
Read and Learn More Glencoe Math Course 2 Volume 1 Common Core Student Edition Solutions
Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 111 Exercise 3 Problem 3
Given:
The table shows the survey of 200 people who have learned to play the instrument in different ways.
Sarah estimates the percentage of people who are self-learned in fractions and in percentages.
To find- Compare the number with the actual number and give
To fill in the table of estimated percent and fraction with the actual percent:
Calculation of percentage:
40%
\(\frac{40}{100}\)= \(\frac{4}{10}\)
= \(\frac{2}{5}\)
30%
\(\frac{30}{100}\)= \(\frac{3}{10}\)
25%
\(\frac{25}{100}\)= \(\frac{1}{4}\)
It is less than the actual number.
This is because we rounding the estimated percent as 25 % from the actual percent 26 %.
So it will cause our estimate to be slightly lower than the actual.
It is less than the actual number. Because we are rounding the percentage, the actual percent becomes slightly less than the estimated percent.
Common Core Chapter 2 Percents Exercise 2.2 Answers Glencoe Math Course 2 Page 114 Exercise 1 Problem 4
Given: 52 % of 10 ≈
To find- Estimate the value
Determine the product by rounding the percentage to the nearest tenth:
52 % of 10 ≈ 50
= \(\frac{1}{2}\)
= 5
Finally, The Value of the estimate is 5.
Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 114 Exercise 3 Problem 5
Given: 151 % of 70 ≈
To find- Estimate the value
Determine the product by rounding the percentage to the nearest tenth:
151 of 70 ≈ 150
⇒ 1.5 × 70
⇒ 105
Finally, The Value of the estimate is 105.
Page 114 Exercise 4 Problem 6
Given: \(\frac{1}{2}\)% of 82 ≈
To find- Estimate the value of the given problem.
Determine the product by rounding the percentage to the nearest tenth:
\(\frac{1}{2}\)% of 82
⇒ \(\frac{1}{2}\)%
= 0.5 %
To find 0.5% of 82
\(\frac{0.5}{100}\) × 82
= 0.005 × 82
= 0.41
≈0.4
Finally, The Value of the estimate is \(\frac{1}{2}\)% of 82 ≈ 0.4
Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 114 Exercise 5 Problem 7
Given: Of the 78 teenagers at a youth camp, 63 have birthdays in the spring.
To find- How many teenagers have birthdays in the spring?
Determine the product by rounding the percentage to the nearest tenth:
63 of 78 ≈ 60
⇒ 0.6 × 78
⇒ 46.8 ≈ 47
As a result, approximately c
Finally, We conclude 47 teenagers celebrate their birthdays in the spring.
Step-By-Step Guide For Exercise 2.2 Chapter 2 Percents In Glencoe Math Course 2 Page 114 Exercise 6 Problem 8
Given: About 0.8 of the land in Maine is federally owned. If Maine has 19,847,680 acres, about how many acres are federally owned? (Example 5)
To find- How many acres are federally owned?
Determine the product by rounding the percentage to the nearest tenth:
0.8% of 19,847,680 ≈
\(\frac{0.8}{100}\) × 19847680
0.008 × 19847680
= 158781.44
Finally, As a result, the feds own approximately 158781 acres.
Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 114 Exercise 7 Problem 9
Given: Estimation of percentage of a number.
Common Method:
Use percent formulas to figure out percentages and unknowns in equations.
Add or subtract a percentage from a number or solve the equations.
There are many formulas for percentage problems. You can think of the most basic as X/Y = P × 100.
The formulas below are all mathematical variations of this formula.
Let’s explore the three basic percentage problems. X and Y are number and P is the percentage:
1. Find P percent of X
2. Find what percent of X is Y.
Example: What is 10% of 150?
Convert the problem to an equation using the percentage formula:
P is 10%, and X is 150, so the equation is 10% × 150 = Y
Convert 10% to a decimal by removing the percent sign and dividing by 100:10/100 = 0.10
Substitute 0.10 for 10% in the equation: 10% × 150 = Y becomes 0.10 × 150 = Y
Do the math: 0.10 × 150 = 15
Y = 15
So 10% of 150 is 15
Double-check your answer with the original question: What is 10% of 150? Multiply 0.10 × 150 = 15
In general, to find n percent of x, we follow these steps:
1. Dividend by 100.
2. Multiply the result by x.
Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 115 Exercise 1 Problem 10
Given:
To convert percentage to a number 47% of 70
Given
47% of 70 ≈ 45
\(\frac{45}{100}\) × 70
= (0.45)70
= 31.5
47% of 70 ≈ 45 = 31.5
The answer for 47 % of 70 ≈ 45 is 31.5
Exercise 2.2 Solutions For Chapter 2 Percents Glencoe Math Course 2 Volume 1 Page 115 Exercise 2 Problem 11
Given:
To convert percentage to a number 39 % of 120
Given
39% of 120 ≈ 40
\(\frac{40}{100}\) × 120
= (0.40)(120)
= 48
39% of 120 ≈ 40 = 48
The answer for 39 % of 120 ≈ 40 = 48
Page 115 Exercise 3 Problem 12
Given:
To convert percentage to a number 21 % of 90
Given
21% of 90 ≈ 20
\(\frac{20}{100}\) × 90
= (0.2)90
= 18
21% of 90 ≈ 20 = 18
The answer for 21 % of 90 ≈ 20 = 18
Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 115 Exercise 4 Problem 13
Given:
To convert percentage to a number 65 % of 152
Given
65 % of 152 ≈ 65
\(\frac{65}{100}\) × 150
= (0.65)150
= 97.5
65 ≈ 65 = 97.5
The answer for 65 % of 152 = 97.5
Common Core Percents Exercise 2.2 Chapter 2 Solutions Glencoe Math Course 2 Page 115 Exercise 5 Problem 14
Given:
To convert percentage to a number 72 % of 238
Given
72 % of 238 ≈ 70
\(\frac{70}{100}\) × 238
= (0.70)238
= 166.6
72 % of 238 ≈ 70 = 166.6
The answer 72 % of 238 = 166.6
Page 115 Exercise 6 Problem 15
Given:
To convert percentage to a number 132% of 54
Given
132 % of 54 ≈ 130
\(\frac{70}{100}\) × 54
= (1.3)54
= 70.2
132 % of 54 ≈130=70.2
The answer 132 % of 54 ≈ 130 = 70.2
Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 115 Exercise 8 Problem 16
Given:
To estimate \(\frac{3}{4}\) % of 168
Given
\(\frac{3}{4}\)% of 168
\(\frac{3}{4}\)% × 168
\(\frac{0.75}{100}\) × 168
= 0.0075 × 168
= 1.26
Therefore, the percentage \(\frac{3}{4}\) of 168 is 1.26.
\(\frac{3}{4}\)% of 168 ≈ 1.3
Examples of problems from Exercise 2.2 Chapter 2 Percents in Glencoe Math Course 2 Page 115 Exercise 9 Problem 17
Given:
To estimate 0.4 % of 510
Given
0.4% of 510
= \(\frac{0.4}{100}\) × 510
= 0.004 × 510
= 2.04
The percentage of 0.4 of 510 is 2.04
Page 115 Exercise 10 Problem 18
Given:
The Financial Literacy Carlie spent $42 at the salon.
Her mother loaned her the money.
Carlie will pay her mother 15% of $42 each week until the loan is repaid.
About how much will Carlie pay each week?
The amount Carlie pay each week is 15 % of $42
= \(\frac{15}{100}\) × 42
= 0.15 × 42
= 6.3
Carlie will pay $6.3 amount each week to her mother until the loan is repaid.
Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 115 Exercise 11 Problem 19
Given:
The United States has 12,383 miles of coastline.
If 0.8 % of the coastline is located in Georgia, about how many miles of coastline are in Georgia?
Given
0.8 % of the coastline in Georgia and 12,383 miles of coastline in United States is
= \(\frac{0.8}{100}\) × 12,383
= 0.008 × 12,383
= 99.064
Approximately 99 miles of coastlines are in Georgia.
Student Edition Glencoe Math Course 2 Chapter 2 Percents Exercise 2.2 Guide Page 116 Exercise 14 Problem 20
Given:
Estimate 54% of 76.8 =?
Given
54% of 76.8 = 50% of 76.8
= \(\frac{50}{100}\) × 76.8
= \(\frac{1}{2}\)×76.8
= 38.4
So,54% of 76.8 is approximately 38.4
The percentage 54 % of 76.8is approximately 38.4
Page 116 Exercise 15 Problem 21
Given:
Estimate 10.5% of 238 =?
Given
10.5% of 238 = \(\frac{105}{1000}\) × 238
= \(\frac{21}{200}\) × 238
= \(\frac{21}{100}\) × 119
= \(\frac{2499}{100}\)
≈ 24.99
So,10.5% of 238 is approximately 24
The percentage of 10.5% of 238 is approximately 24
Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 116 Exercise 16 Problem 22
Given:
The average white rhinoceros gives birth to a single calf that weight about 3.8% as much as its mother rhinoceros weight 3.75 tons, about how many pounds does its calf weight?
Given
3.8% of 3.75 = \(\frac{3.8}{100}\) × 3.75
= \(\frac{380}{10000}\) × 375
= \(\frac{19}{500}\) × 375
= \(\frac{19}{500}\) × 75
= 14.25t
≈ 0.145t (In pounds)
So, the baby animal weights in 0.145t
Step-by-step answers for Exercise 2.2 Chapter 2 Percents in Glencoe Math Course 2 Volume 1 Page 116 Exercise 18 Problem 23
Given:
Explain how you could find % of $800
By simplifying fraction to omit % symbol and multiplying the values.
\(\frac{3}{8}\)% of 800 = \(\frac{3}{8}\)×800×\(\frac{1}{100}\)
= 3
\(\frac{3}{8}\)% of 800 = 3
The\(\frac{3}{8}\)% answer of 800 is 3.
Page 116 Exercise 19 Problem 24
Is an estimate for the percent of a number always sometimes or never greater than the actual percent of the number?
Give an example or a counterexample to support your answer
An estimate for the percent of a number is sometimes greater than the actual percent of the number
One estimate for 18% of 40 is, \(\frac{1}{5}\).40 = 8
While ,one estimate for 22% of 60 is ,\(\frac{1}{5}\).60 = 12
While never greater than the actual percent,50% of 30 is
\(\frac{1}{2}\).30 = 15
It is the example for the percent of a number sometimes, or never greater than the actual percent of the number
Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 116 Exercise 20 Problem 25
Given:
Cost of bedroom furniture=$1,789.43
Percentage cost of dresser=39.7 of total cost
To find- Cost of the dresser?
Cost of dresser
= 39.7% of $1,789.43
= (\(\frac{40}{100}\) × 1,789.43) − (\(\frac{0.3}{100}\) × 1,789.43)
= 715.772 − 5.368
= 710.404 ≈ $720
Hence, $720 is the best estimate for the cost of the dresser.
Page 117 Exercise 21 Problem 26
Given:
76%of 180 ≈ ?
To find- Evaluate the problem.
76% 180 = 75% of 180 + 1% of 180
= (\(\frac{75}{100}\) × 180) + (\(\frac{1}{100}\) × 180)
= 135 + 1.8
= 136.8 ≈ 137
Therefore by evaluating the equation the percentage of 76% of 180 ≈ 137
Page 117 Exercise 22 Problem 27
Given:
57%of 29 ≈?
To find- Evaluate the problem.
57% of 29
= (\(\frac{60}{100}\) × 29) − (\(\frac{3}{100}\) × 29 )
= 17.4 − 0.87
= 16.53 ≈ 17
Therefore by evaluating the equation the percentage of 57% of 29 ≈ 17
Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 117 Exercise 23 Problem 28
Given:
92%of 104 ≈ ?
To find- Evaluate the problem.
Let, 92% of 104
= (\(\frac{100}{100} × 104\)) − (\(\frac{8}{100} × 104\))
= 104 − 8.32
= 95.68 ≈ 96
Therefore by evaluating the equation the percentage of 92 of 104 ≈ 96
Page 117 Exercise 25 Problem 29
Given:
0.9% of 74 ≈ ?
To find- Evaluate the problem.
0.9% = (1−0.1)%
74 × \(\frac{1-0.1}{100}\) = \(\frac{74×1}{100}\)– \(\frac{74×0.1}{100}\)
= \(\frac{74}{100}\)–\(\frac{7.4}{100}\)
= 0.74 − 0.074
= 0.666
Therefore by evaluating the equation the percentage of 0.9 of 74 = 0.666
Page 117 Exercise 26 Problem 30
Given:
32% of 89.9 ≈ ?
To find- Evaluate the problem.
30%of89.9 = \(\frac{30}{100}\) × 89.9
= 26.97
2% of 89.9 = \(\frac{2}{100}\) × 89.9
= 1.798
32 % of 89.9 = 26.97 + 1.798 = 28.768
Therefore by evaluating the equation the percentage of 32 % of 89.9 = 28.8
Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 117 Exercise 27 Problem 31
Given:
Total muscles to frown = 43
Percentage of muscles used to smile = 32%
To find- A number of muscles used to smile?
Number of muscles used to smile
= 32%of43
= (\(\frac{30}{100}\)) × 43 + (\(\frac{32}{100}\)) × 43
= 12.9 + 0.86
= 13.76 ≈ 14
Therefore,14 muscles are used when using a smile.
Page 117 Exercise 28 Problem 32
Given:
Coastline of Atlantic coast = 2.069miles
Percentage of coastlines in New Hampshire = \(\frac{6}{10}\)%
To find- Length of coastlines in New Hampshire?
Length of coastlines in New Hampshire = \(\frac{6}{10}\)% of 2.069
\(\frac{0.6}{10}\)% × 2.069 = 0.0124
Therefore, the length of coastlines in New Hampshire is 0.0124 miles.
Page 118 Exercise 32 Problem 33
Given:
5n = 120
To find- The value of n =?
The value of n is
5n = 120
n = 120/5
= 24
Therefore, the value of n is 24.
Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 118 Exercise 33 Problem 34
Given:
1,200 = 4a
To find – Solve each equation show your work
The given equation is 1200 = 4a
Divide both sides by ‘4’, and we get
\(\frac{1200}{4}\) = \(\frac{44}{4}\)
a = 300
By solving the given equation we get a = 300
Page 118 Exercise 34 Problem 35
Given:
6x = 39
To find – Solve each equation show your work
The given equation is 6x = 39
Divide both sides by ‘6’, and we get
\(\frac{6x}{6}\) = \(\frac{39}{6}\)
x = 6.5
By solving the given equation we get x = 6.5
Page 118 Exercise 36 Problem 36
Given:\(\frac{3}{5}\)
To find – Write three fraction equation
By multiplying the numerator and the denominator by 2,3and 4we get
The equivalent fractional to \(\frac{3}{5}\) is
\(\frac{6}{10}\), \(\frac{9}{15}\), and \(\frac{12}{20}\)
Three fractions equivalent to \(\frac{3}{5}\), \(\frac{6}{10}\),\(\frac{9}{15}\) and \(\frac{12}{20}\)
Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 120 Exercise 1 Problem 37
Given:
The bar diagram for eighth and tenth grade.
To find-
Total tickets sold above each bar. We will divide the bar into ten equal sections.
Each section represents ten percent.
Let us divide the bar into 10 equal sections.
The bar for eighth and seventh grade are similar.
By measuring the above bar, we found that 50 % of tickets were sold.
The bar diagrams below show 100% for each grade. Abel Divide each bar into 10 equal sections.
So, each section will represent 10%. The total number of tickets to be sold above each bar is 50 %.
Page 120 Exercise 2 Problem 38
Given:
To Find -The number that belongs in each section.
Then write that Section.
Calculation:
300 ÷ 10 = 30
250 ÷ 10 = 25
The number of tickets in each section of eight & seventh-grade baris 30 and 25 respectively.
Page 120 Exercise 3 Problem 39
Given:
To find- The number of sections to shade for each bar.
Then shade the sections.
Eight
225 ÷ 30 = 7.5
Seventh
200 ÷ 25 = 8
The number of sections to be Shaded in the eighth and seventh-grade bar is 7.5 and 8 respectively.
The eighth grade sold 75 % of their tickets. The seventh grade sold 80% of their tickets.
The Seventh grade sold the greater percent of their tickets.
Hence, the Seventh grade sold a greater percent of their tickets